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Shared Qs (020)


  1. Question

    Solve:

    \[6 = 3^x \]

    Tolerance is \(\pm0.01\).


    Solution


  2. Question

    Solve:

    \[10 = 9^x \]

    Tolerance is \(\pm0.01\).


    Solution


  3. Question

    Solve:

    \[20 = 2^x \]

    Tolerance is \(\pm0.01\).


    Solution


  4. Question

    Solve:

    \[26 = 2^x \]

    Tolerance is \(\pm0.01\).


    Solution


  5. Question

    Solve:

    \[2 = 6^x \]

    Tolerance is \(\pm0.01\).


    Solution


  6. Question

    Solve:

    \[4 = 9^x \]

    Tolerance is \(\pm0.01\).


    Solution


  7. Question

    Solve:

    \[18 = 2^x \]

    Tolerance is \(\pm0.01\).


    Solution


  8. Question

    Solve:

    \[15 = 9^x \]

    Tolerance is \(\pm0.01\).


    Solution


  9. Question

    Solve:

    \[29 = 4^x \]

    Tolerance is \(\pm0.01\).


    Solution


  10. Question

    Solve:

    \[34 = 9^x \]

    Tolerance is \(\pm0.01\).


    Solution


  11. Question

    Let’s say a warm block initially at 74.5°C is placed outside on a moderately windy day. The outdoor temperature is 14.9°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.08 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.08(T-14.9)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 8.664 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 59.6e^{-0.08 t}+14.9\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 22.1 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  12. Question

    Let’s say a warm block initially at 62.9°C is placed outside on a moderately windy day. The outdoor temperature is 15°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.08 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.08(T-15)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 8.664 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 47.9e^{-0.08 t}+15\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 29.2 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  13. Question

    Let’s say a warm block initially at 84.5°C is placed outside on a moderately windy day. The outdoor temperature is 12.2°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.32 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.32(T-12.2)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 2.166 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 72.3e^{-0.32 t}+12.2\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 27.6 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  14. Question

    Let’s say a warm block initially at 58.1°C is placed outside on a moderately windy day. The outdoor temperature is 24.4°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.07 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.07(T-24.4)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 9.902 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 33.7e^{-0.07 t}+24.4\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 35 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  15. Question

    Let’s say a warm block initially at 97.5°C is placed outside on a moderately windy day. The outdoor temperature is 5.6°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.24 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.24(T-5.6)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 2.888 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 91.9e^{-0.24 t}+5.6\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 14.8 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  16. Question

    Let’s say a warm block initially at 100.7°C is placed outside on a moderately windy day. The outdoor temperature is 25.6°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.06 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.06(T-25.6)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 11.552 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 75.1e^{-0.06 t}+25.6\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 37.9 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  17. Question

    Let’s say a warm block initially at 96.7°C is placed outside on a moderately windy day. The outdoor temperature is 26.2°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.32 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.32(T-26.2)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 2.166 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 70.5e^{-0.32 t}+26.2\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 58.7 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  18. Question

    Let’s say a warm block initially at 55.5°C is placed outside on a moderately windy day. The outdoor temperature is 4.8°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.32 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.32(T-4.8)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 2.166 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 50.7e^{-0.32 t}+4.8\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 18.8 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  19. Question

    Let’s say a warm block initially at 79.8°C is placed outside on a moderately windy day. The outdoor temperature is 18.2°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.09 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.09(T-18.2)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 7.702 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 61.6e^{-0.09 t}+18.2\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 48.2 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  20. Question

    Let’s say a warm block initially at 103.9°C is placed outside on a moderately windy day. The outdoor temperature is 17°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:

    For this block, all the effects combine into a single proportionality constant: 0.04 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.04(T-17)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 17.329 minutes, the temperature difference halves.

    By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.

    \[T(t) ~=~ 86.9e^{-0.04 t}+17\]

    where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 37.7 degrees Celsius?

    The tolerance is \(\pm0.01\) min.


    Solution


  21. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 22°C.

    \[T(0) = 22 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.44 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 13°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.44 T+13\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.44T+13)\cdot\Delta t\)
    0 22 0.1 0.332
    0.1 22.332 0.1 0.317392
    0.2 22.649392 0.1 0.3034268
    0.3 22.9528188 0.1 0.290076

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 22+0.332 = 22.332\).


    Solution


  22. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 25°C.

    \[T(0) = 25 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.29 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 14°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.29 T+14\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.29T+14)\cdot\Delta t\)
    0 25 0.1 0.675
    0.1 25.675 0.1 0.655425
    0.2 26.330425 0.1 0.6364177
    0.3 26.9668427 0.1 0.6179616

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 25+0.675 = 25.675\).


    Solution


  23. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 18°C.

    \[T(0) = 18 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.28 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 9°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.28 T+9\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.28T+9)\cdot\Delta t\)
    0 18 0.1 0.396
    0.1 18.396 0.1 0.384912
    0.2 18.780912 0.1 0.3741345
    0.3 19.1550465 0.1 0.3636587

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 18+0.396 = 18.396\).


    Solution


  24. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 26°C.

    \[T(0) = 26 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.49 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 15°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.49 T+15\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.49T+15)\cdot\Delta t\)
    0 26 0.1 0.226
    0.1 26.226 0.1 0.214926
    0.2 26.440926 0.1 0.2043946
    0.3 26.6453206 0.1 0.1943793

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 26+0.226 = 26.226\).


    Solution


  25. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 23°C.

    \[T(0) = 23 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.26 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 12°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.26 T+12\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.26T+12)\cdot\Delta t\)
    0 23 0.1 0.602
    0.1 23.602 0.1 0.586348
    0.2 24.188348 0.1 0.571103
    0.3 24.759451 0.1 0.5562543

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 23+0.602 = 23.602\).


    Solution


  26. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 28°C.

    \[T(0) = 28 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.3 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 16°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.3 T+16\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.3T+16)\cdot\Delta t\)
    0 28 0.1 0.76
    0.1 28.76 0.1 0.7372
    0.2 29.4972 0.1 0.715084
    0.3 30.212284 0.1 0.6936315

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 28+0.76 = 28.76\).


    Solution


  27. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 18°C.

    \[T(0) = 18 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.14 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 7°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.14 T+7\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.14T+7)\cdot\Delta t\)
    0 18 0.1 0.448
    0.1 18.448 0.1 0.441728
    0.2 18.889728 0.1 0.4355438
    0.3 19.3252718 0.1 0.4294462

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 18+0.448 = 18.448\).


    Solution


  28. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 13°C.

    \[T(0) = 13 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.43 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 8°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.43 T+8\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.43T+8)\cdot\Delta t\)
    0 13 0.1 0.241
    0.1 13.241 0.1 0.230637
    0.2 13.471637 0.1 0.2207196
    0.3 13.6923566 0.1 0.2112287

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 13+0.241 = 13.241\).


    Solution


  29. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 29°C.

    \[T(0) = 29 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.37 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 15°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.37 T+15\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.37T+15)\cdot\Delta t\)
    0 29 0.1 0.427
    0.1 29.427 0.1 0.411201
    0.2 29.838201 0.1 0.3959866
    0.3 30.2341876 0.1 0.3813351

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 29+0.427 = 29.427\).


    Solution


  30. Question

    Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.

    Let’s say that when \(t=0\) min, the water temperature is 23°C.

    \[T(0) = 23 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.48 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 15°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).

    \[\frac{\Delta T}{\Delta t} ~\approx~ -0.48 T+15\]

    Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.

    \(t\) \(T\) \(\Delta t\) \(\Delta T ~=~ (-0.48T+15)\cdot\Delta t\)
    0 23 0.1 0.396
    0.1 23.396 0.1 0.376992
    0.2 23.772992 0.1 0.3588964
    0.3 24.1318884 0.1 0.3416694

    Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 23+0.396 = 23.396\).


    Solution


  31. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.025v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 700 0.1 -0.981 0
    0.1 -0.981 700 0.1 -0.9785941 -0.0981
    0.2 -1.9595941 699.9019 0.1 -0.9714 -0.1959594
    0.3 -2.9309941 699.7059406 0.1 -0.9595232 -0.2930994
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  32. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.026v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 500 0.1 -0.981 0
    0.1 -0.981 500 0.1 -0.9784979 -0.0981
    0.2 -1.9594979 499.9019 0.1 -0.971017 -0.1959498
    0.3 -2.9305148 499.7059502 0.1 -0.9586714 -0.2930515
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  33. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.04v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 700 0.1 -0.981 0
    0.1 -0.981 700 0.1 -0.9771506 -0.0981
    0.2 -1.9581506 699.9019 0.1 -0.9656626 -0.1958151
    0.3 -2.9238131 699.7060849 0.1 -0.9468053 -0.2923813
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  34. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.027v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 700 0.1 -0.981 0
    0.1 -0.981 700 0.1 -0.9784016 -0.0981
    0.2 -1.9594016 699.9019 0.1 -0.970634 -0.1959402
    0.3 -2.9300356 699.7059598 0.1 -0.9578202 -0.2930036
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  35. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.048v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 700 0.1 -0.981 0
    0.1 -0.981 700 0.1 -0.9763807 -0.0981
    0.2 -1.9573807 699.9019 0.1 -0.9626096 -0.1957381
    0.3 -2.9199902 699.7061619 0.1 -0.9400736 -0.291999
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  36. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.043v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 400 0.1 -0.981 0
    0.1 -0.981 400 0.1 -0.9768618 -0.0981
    0.2 -1.9578618 399.9019 0.1 -0.9645171 -0.1957862
    0.3 -2.922379 399.7061138 0.1 -0.9442767 -0.2922379
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  37. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.033v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 300 0.1 -0.981 0
    0.1 -0.981 300 0.1 -0.9778242 -0.0981
    0.2 -1.9588242 299.9019 0.1 -0.9683379 -0.1958824
    0.3 -2.9271621 299.7060176 0.1 -0.9527247 -0.2927162
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  38. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.045v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 600 0.1 -0.981 0
    0.1 -0.981 600 0.1 -0.9766694 -0.0981
    0.2 -1.9576694 599.9019 0.1 -0.9637539 -0.1957669
    0.3 -2.9214233 599.7061331 0.1 -0.9425938 -0.2921423
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  39. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.037v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 300 0.1 -0.981 0
    0.1 -0.981 300 0.1 -0.9774393 -0.0981
    0.2 -1.9584393 299.9019 0.1 -0.9668087 -0.1958439
    0.3 -2.925248 299.7060561 0.1 -0.9493388 -0.2925248
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  40. Question

    When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.

    \[\frac{\Delta v}{\Delta t} = -g+kv^2\]

    The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.

    Because the object is falling straight down, velocity is how fast the height is changing.

    \[\frac{\Delta h}{\Delta t} = v\]

    Let’s set the parameters and initial values.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(v\) \(h\) \(\Delta t\) \(\Delta v ~=~ (-9.81+0.031v^2)\cdot\Delta t\) \(\Delta h ~=~ v \cdot \Delta t\)
    0 0 500 0.1 -0.981 0
    0.1 -0.981 500 0.1 -0.9780167 -0.0981
    0.2 -1.9590167 499.9019 0.1 -0.969103 -0.1959017
    0.3 -2.9281197 499.7059983 0.1 -0.954421 -0.292812
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    The tolerance is \(\pm 0.01\) m.


    Solution


  41. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.024\cdot x\cdot (1-x/13)\cdot\Delta t\)
    0 1.5 2 0.0636923
    2 1.5636923 2 0.066029
    4 1.6297214 2 0.0684199
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  42. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.023\cdot x\cdot (1-x/9)\cdot\Delta t\)
    0 1.1 2 0.0444156
    2 1.1444156 2 0.0459492
    4 1.1903647 2 0.0475145
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  43. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.019\cdot x\cdot (1-x/19)\cdot\Delta t\)
    0 1.9 2 0.06498
    2 1.96498 2 0.0669469
    4 2.0319269 2 0.0689558
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  44. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.02\cdot x\cdot (1-x/7)\cdot\Delta t\)
    0 1.1 2 0.0370857
    2 1.1370857 2 0.0380951
    4 1.1751808 2 0.0391155
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  45. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.02\cdot x\cdot (1-x/18)\cdot\Delta t\)
    0 1.7 2 0.0615778
    2 1.7615778 2 0.0635672
    4 1.825145 2 0.0656032
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  46. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.012\cdot x\cdot (1-x/17)\cdot\Delta t\)
    0 2.9 2 0.0577271
    2 2.9577271 2 0.0586351
    4 3.0163622 2 0.0595478
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  47. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.012\cdot x\cdot (1-x/15)\cdot\Delta t\)
    0 2.6 2 0.051584
    2 2.651584 2 0.0523886
    4 2.7039726 2 0.053197
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  48. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.01\cdot x\cdot (1-x/11)\cdot\Delta t\)
    0 1.8 2 0.0301091
    2 1.8301091 2 0.0305125
    4 1.8606216 2 0.030918
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  49. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.022\cdot x\cdot (1-x/8)\cdot\Delta t\)
    0 1.9 2 0.063745
    2 1.963745 2 0.0651952
    4 2.0289402 2 0.0666321
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  50. Question

    The competitive Lotka–Volterra equation is used to simulate population growth.

    \[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]

    where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.

    For our simulation, use the following values for the initial values and parameters.

    Using Euler’s method, we can produce the first few rounds of the simulation.

    \(t\) \(x\) \(\Delta t\) \(\Delta x ~=~ 0.015\cdot x\cdot (1-x/8)\cdot\Delta t\)
    0 2.5 2 0.0515625
    2 2.5515625 2 0.0521326
    4 2.6036951 2 0.0526887
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)

    Solution


  51. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.7 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.5c\]

    Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.


    Solution


  52. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.6c\]

    Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.


    Solution


  53. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.5 b-0.2bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.9c\]

    Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.


    Solution


  54. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.7 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.9c\]

    Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.


    Solution


  55. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.9c\]

    Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.


    Solution


  56. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.8c\]

    Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.


    Solution


  57. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.9c\]

    Use Euler’s method to estimate the number of coyotes (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand coyotes.


    Solution


  58. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.4 b-0.2bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.5c\]

    Use Euler’s method to estimate the number of coyotes (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand coyotes.


    Solution


  59. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.6c\]

    Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.


    Solution


  60. Question

    The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.

    Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.

    \[\frac{\Delta b}{\Delta t} = 1.6 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.7c\]

    Use Euler’s method to estimate the number of coyotes (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand coyotes.


    Solution


  61. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = 2.1\] \[y_0 = 0.8\] \[z_0 = -1.1\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  62. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.7\] \[y_0 = 1.5\] \[z_0 = -0.2\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  63. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.6\] \[y_0 = -2.4\] \[z_0 = 0.6\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  64. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.8\] \[y_0 = 2.7\] \[z_0 = -3\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  65. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.5\] \[y_0 = 1.1\] \[z_0 = -2.7\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  66. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = 0.8\] \[y_0 = 1.6\] \[z_0 = 2.4\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  67. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = 1.3\] \[y_0 = 0.1\] \[z_0 = 2\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  68. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.8\] \[y_0 = -1\] \[z_0 = 1.4\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  69. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = -2.3\] \[y_0 = 1.7\] \[z_0 = -0.6\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  70. Question

    In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).

    \[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]

    Use these initial conditions: \[t_0 = 0\] \[x_0 = 2\] \[y_0 = -1.1\] \[z_0 = -0.7\]

    Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E
    6. F

    Solution


  71. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.25\sin(4.46z+3.96)+0.48\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:

    \(z\) \(r=0.25\cdot\sin(4.46z+3.96)+0.48\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.2974854 0 0.02 0.0111209
    0.02 0.2829917 0.0111209 0.02 0.0100637
    0.04 0.2700646 0.0211846 0.02 0.0091653
    0.06 0.2588067 0.0303499 0.02 0.0084171
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.6910107 ??? 0.02

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  72. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.26\sin(5.61z+1.49)+0.48\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:

    \(z\) \(r=0.26\cdot\sin(5.61z+1.49)+0.48\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.7391518 0 0.02 0.0686558
    0.02 0.7398718 0.0686558 0.02 0.0687896
    0.04 0.7373237 0.1374454 0.02 0.0683166
    0.06 0.7315397 0.205762 0.02 0.067249
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.669532 ??? 0.02

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  73. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.3\sin(4.26z+5.12)+0.51\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:

    \(z\) \(r=0.3\cdot\sin(4.26z+5.12)+0.51\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.2345789 0 0.02 0.0069149
    0.02 0.2456981 0.0069149 0.02 0.007586
    0.04 0.2587347 0.0145009 0.02 0.0084124
    0.06 0.2735942 0.0229133 0.02 0.0094064
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.5234289 ??? 0.02

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  74. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.12\sin(5.23z+2.82)+0.48\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:

    \(z\) \(r=0.12\cdot\sin(5.23z+2.82)+0.48\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.5179294 0 0.02 0.0337094
    0.02 0.5058353 0.0337094 0.02 0.0321535
    0.04 0.4934587 0.0658629 0.02 0.0305993
    0.06 0.4809351 0.0964622 0.02 0.0290658
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.597702 ??? 0.02

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  75. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.16\sin(3.38z+2.85)+0.49\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.05\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.05\). Your table should look like the following:

    \(z\) \(r=0.16\cdot\sin(3.38z+2.85)+0.49\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.5359965 0 0.05 0.0902555
    0.05 0.5095657 0.0902555 0.05 0.0815737
    0.1 0.4825775 0.1718292 0.05 0.0731617
    0.15 0.4558007 0.244991 0.05 0.0652679
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.4814944 ??? 0.05

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  76. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.16\sin(5.29z+2.28)+0.49\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:

    \(z\) \(r=0.16\cdot\sin(5.29z+2.28)+0.49\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.6114209 0 0.02 0.0469776
    0.02 0.5997385 0.0469776 0.02 0.0451995
    0.04 0.5868289 0.0921771 0.02 0.0432746
    0.06 0.5728364 0.1354516 0.02 0.0412355
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.6435916 ??? 0.02

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  77. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.36\sin(4.1z+0.11)+0.51\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.04\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.04\). Your table should look like the following:

    \(z\) \(r=0.36\cdot\sin(4.1z+0.11)+0.51\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.5495202 0 0.04 0.075894
    0.04 0.6074104 0.075894 0.04 0.0927266
    0.08 0.6626865 0.1686205 0.04 0.1103713
    0.12 0.7138651 0.2789918 0.04 0.1280773
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.1944835 ??? 0.04

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  78. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.21\sin(3.9z+0.17)+0.54\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.05\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.05\). Your table should look like the following:

    \(z\) \(r=0.21\cdot\sin(3.9z+0.17)+0.54\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.5755283 0 0.05 0.1040599
    0.05 0.6149594 0.1040599 0.05 0.1188072
    0.1 0.6515491 0.222867 0.05 0.1333657
    0.15 0.6839106 0.3562327 0.05 0.1469429
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.37186 ??? 0.05

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  79. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.18\sin(6.93z+5.39)+0.5\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.04\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.04\). Your table should look like the following:

    \(z\) \(r=0.18\cdot\sin(6.93z+5.39)+0.5\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.3597669 0 0.04 0.0325299
    0.04 0.3960027 0.0325299 0.04 0.0394127
    0.08 0.4401785 0.0719426 0.04 0.0486965
    0.12 0.4889217 0.120639 0.04 0.0600784
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.4561006 ??? 0.04

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  80. Question

    A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.

    \[r ~=~ 0.13\sin(7.22z+0.62)+0.5\]

    plot of chunk unnamed-chunk-2

    To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.04\) for each disc.

    plot of chunk unnamed-chunk-3

    Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.

    plot of chunk unnamed-chunk-4

    \[\Delta V = \pi r^2 \cdot \Delta z\]

    Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.04\). Your table should look like the following:

    \(z\) \(r=0.13\cdot\sin(7.22z+0.62)+0.5\) \(V\) \(\Delta z\) \(\Delta V = 2\pi r^2\cdot\Delta z\)
    0 0.5755346 0 0.04 0.0832497
    0.04 0.6025397 0.0832497 0.04 0.0912454
    0.08 0.6210517 0.1744951 0.04 0.0969383
    0.12 0.6295373 0.2714334 0.04 0.0996054
    \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\) \(\vdots\)
    1 0.6299873 ??? 0.04

    The tolerance is \(\pm 0.01\) cubic meters.


    Solution


  81. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = 0.1\] \[y_0 = -7.5\] \[z_0 = 28.2\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  82. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = 14.5\] \[y_0 = -2\] \[z_0 = 26.7\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  83. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = 1.8\] \[y_0 = -9.3\] \[z_0 = 29.1\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  84. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = -4.2\] \[y_0 = -1.6\] \[z_0 = 32.8\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  85. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = -11.6\] \[y_0 = -7.7\] \[z_0 = 23.2\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  86. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = 0.6\] \[y_0 = 7.6\] \[z_0 = 33.9\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  87. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = -10.6\] \[y_0 = -0.5\] \[z_0 = 35.1\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  88. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = -7.5\] \[y_0 = -9.6\] \[z_0 = 35.9\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  89. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = 4.4\] \[y_0 = 7\] \[z_0 = 37.1\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  90. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:

    \[x_0 = -11.2\] \[y_0 = 5.1\] \[z_0 = 24.9\]

    Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  91. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = -10.4\] \[y_0 = -4.2\] \[z_0 = 38.6\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  92. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = -4.7\] \[y_0 = -5.6\] \[z_0 = 29.4\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  93. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = -0.5\] \[y_0 = 5.6\] \[z_0 = 26.5\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  94. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = -9.7\] \[y_0 = 1.7\] \[z_0 = 24.1\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  95. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = 2.1\] \[y_0 = 5.7\] \[z_0 = 24.6\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  96. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = 10.3\] \[y_0 = 0.8\] \[z_0 = 24.3\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  97. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = 11.5\] \[y_0 = 5.3\] \[z_0 = 16.1\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  98. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = 10.1\] \[y_0 = -0.5\] \[z_0 = 21.4\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  99. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = -11.9\] \[y_0 = -0.4\] \[z_0 = 24.8\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  100. Question

    By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.

    \[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]

    Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.

    \[x_0 = 8.4\] \[y_0 = -0.9\] \[z_0 = 34.2\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.

    plot of chunk unnamed-chunk-2


    1. A
    2. B
    3. C
    4. D
    5. E

    Solution


  101. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  102. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  103. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  104. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  105. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  106. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  107. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  108. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  109. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  110. Question

    The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.

    plot of chunk unnamed-chunk-2

    What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  111. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  112. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  113. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  114. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  115. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  116. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  117. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  118. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  119. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  120. Question

    The doubling time of an exponentially growing function is the amount of time it takes for the population to double.

    plot of chunk unnamed-chunk-2

    What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)


    Solution


  121. Question

    At \(t=0\) a population is \(y(0)=7.5\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.11 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 7.5e^{0.11t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  122. Question

    At \(t=0\) a population is \(y(0)=6.9\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.09 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 6.9e^{0.09t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  123. Question

    At \(t=0\) a population is \(y(0)=4.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.3 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 4.7e^{0.3t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  124. Question

    At \(t=0\) a population is \(y(0)=7.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.1 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 7.7e^{0.1t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  125. Question

    At \(t=0\) a population is \(y(0)=7.5\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.21 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 7.5e^{0.21t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  126. Question

    At \(t=0\) a population is \(y(0)=6.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.11 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 6.7e^{0.11t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  127. Question

    At \(t=0\) a population is \(y(0)=12.5\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.15 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 12.5e^{0.15t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  128. Question

    At \(t=0\) a population is \(y(0)=15.9\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.09 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 15.9e^{0.09t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  129. Question

    At \(t=0\) a population is \(y(0)=14.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.12 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 14.7e^{0.12t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  130. Question

    At \(t=0\) a population is \(y(0)=16.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.26 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 16.7e^{0.26t}\]

    What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  131. Question

    At \(t=0\) a population is \(y(0)=5.9\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.12 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 5.9e^{-0.12t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  132. Question

    At \(t=0\) a population is \(y(0)=11.9\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.18 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 11.9e^{-0.18t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  133. Question

    At \(t=0\) a population is \(y(0)=8.2\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.17 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 8.2e^{-0.17t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  134. Question

    At \(t=0\) a population is \(y(0)=15.1\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.18 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 15.1e^{-0.18t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  135. Question

    At \(t=0\) a population is \(y(0)=7.8\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.25 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 7.8e^{-0.25t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  136. Question

    At \(t=0\) a population is \(y(0)=10.9\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.09 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 10.9e^{-0.09t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  137. Question

    At \(t=0\) a population is \(y(0)=12.5\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.14 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 12.5e^{-0.14t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  138. Question

    At \(t=0\) a population is \(y(0)=6\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.33 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 6e^{-0.33t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  139. Question

    At \(t=0\) a population is \(y(0)=6.5\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.09 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 6.5e^{-0.09t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  140. Question

    At \(t=0\) a population is \(y(0)=18.8\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.16 y\]

    where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.

    \[y = 18.8e^{-0.16t}\]

    What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)


    Solution


  141. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(6\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.81^{x}\\\\ g(x) &= 6\cdot 1.81^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.81^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  142. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(4\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.78^{x}\\\\ g(x) &= 4\cdot 1.78^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.78^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  143. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(6\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.87^{x}\\\\ g(x) &= 6\cdot 1.87^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.87^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  144. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(4\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.5^{x}\\\\ g(x) &= 4\cdot 1.5^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.5^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  145. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(3\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.22^{x}\\\\ g(x) &= 3\cdot 1.22^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.22^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  146. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(3\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.31^{x}\\\\ g(x) &= 3\cdot 1.31^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.31^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  147. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(5\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.32^{x}\\\\ g(x) &= 5\cdot 1.32^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.32^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  148. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(5\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.27^{x}\\\\ g(x) &= 5\cdot 1.27^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.27^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  149. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(5\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.48^{x}\\\\ g(x) &= 5\cdot 1.48^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.48^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  150. Question

    For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(4\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= 1.41^{x}\\\\ g(x) &= 4\cdot 1.41^{x} \end{align}\]

    Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).

    \[g(x) = f(x+h)\]

    Or,

    \[g(x) = 1.41^{x+h}\]

    where \(h\) represents the distance of the horizontal shift.


    Solution


  151. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(3.1\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/3.1} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  152. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-4.5\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/4.5} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  153. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-4.5\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/4.5} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  154. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-3.4\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/3.4} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  155. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(4.3\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/4.3} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  156. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-4.1\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/4.1} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  157. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-4.2\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/4.2} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  158. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(3.1\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/3.1} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  159. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(2.1\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/2.1} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution


  160. Question

    For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(4.4\):

    plot of chunk unnamed-chunk-2

    \[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/4.4} \end{align}\]

    Instead of using the horizontal stretch transformation, we could have changed the base.

    \[g(x) = B^{x}\]

    where \(B\) represents the new base.


    Solution